Order $6$ has exactly two group types: cyclic and dihedral.Suggest a slogan
Up to isomorphism, the groups of order \(6\) are the cyclic group \(\ZZ/6\ZZ\), which is abelian, and the dihedral group \(D_3\cong S_3\) of order \(6\), which is nonabelian.
Let \(\abs G=6\). By Sylow's theorems, the number of Sylow \(3\)-subgroups divides \(2\) and is congruent to \(1\) modulo \(3\), so there is a unique, hence normal, Sylow \(3\)-subgroup \(N\cong\ZZ/3\ZZ\). A Sylow \(2\)-subgroup \(H\cong\ZZ/2\ZZ\) meets \(N\) trivially and \(NH=G\), so \(G\cong N\rtimes_\theta H\) for a homomorphism \(\theta\colon H\to\Aut(N)\cong\theset{\pm1}\). If \(\theta\) is trivial, \(G\cong\ZZ/3\ZZ\times\ZZ/2\ZZ\cong\ZZ/6\ZZ\); otherwise the generator of \(H\) acts by inversion and \(G\cong D_3\).
What are the groups of order 6?